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# sampling backgammon dice

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### Chuck Bower

Dec 13, 1999, 3:00:00 AM12/13/99
to
In another thread ("Cooked Dice") there appeared to be a question
about whether the last roll of a recorded game would contain bias, even
if the dice generator itself were unbiased. E.g. is the probability that
a game ends with doubles more than 1/6, less than 1/6, or exactly 1/6?

Consider the following proposition:

A neutral third party will roll a pair of fair dice. Two players take
on the following roles:

If the dice roll comes up doublets, player A will pay player B five units.
If the dice roll comes up NON-doublets, player B will pay player A one unit.
Player A can call the game off at any time (after settling up, of course).
Player B MUST continue to play the game as long as player A wants to.

Assuming both players have sufficient bankrolls that neither is going
to go broke, who has the advantage in this game? Would it matter if player
B were the one who gets to decide when the game ends? What about if the
neutral third party (dice roller) were the one who decided when the game
ends?

Chuck
bo...@bigbang.astro.indiana.edu
c_ray on FIBS

### JP White

Dec 13, 1999, 3:00:00 AM12/13/99
to
Chuck Bower wrote:

> In another thread ("Cooked Dice") there appeared to be a question
> about whether the last roll of a recorded game would contain bias, even
> if the dice generator itself were unbiased. E.g. is the probability that
> a game ends with doubles more than 1/6, less than 1/6, or exactly 1/6?
>

I have pondered this myself for the last few days.

>
> Consider the following proposition:
>
> A neutral third party will roll a pair of fair dice. Two players take
> on the following roles:
>
> If the dice roll comes up doublets, player A will pay player B five units.
> If the dice roll comes up NON-doublets, player B will pay player A one unit.
> Player A can call the game off at any time (after settling up, of course).
> Player B MUST continue to play the game as long as player A wants to.
>
> Assuming both players have sufficient bankrolls that neither is going
> to go broke, who has the advantage in this game?

I imagine A, since he decides when to stop. He'll stop when he's ahead.

> Would it matter if player
> B were the one who gets to decide when the game ends?

Same again. He'll stop when he's ahead.

> neutral third party (dice roller) were the one who decided when the game
> ends?

Then no one has a real advantage (Unless one of them does a deal with the
roller)

My first thought when I read the proposition that more games end in doubles was
'poppycock'.

Then I started to think about it some more. It's not when the double comes along
that's important, it's the effect it has. So I started to see the reasoning, a
double will take more men off at the end of a game and may decide it. Hmmm

Then I thought some more.

Let's say two backgammon players had all their checkers in their respective home
boards in exactly the same configuration. A non contact bearoff.

Let's also assume both players roll identically (say consecutive 6-5's) and are
both able to bearoff two men each roll.

The first player on roll will win on his 8th roll.

If on the 7th roll the second player on roll rolls 6-6 then he wins, before the
first player can bearoff his last checker.

So far so good, theory is holding up.

But upon further analysis, it is clear that it doesn't matter *when* the 6-6 is
rolled by the second player. 1st roll, 2nd roll, 3rd roll..the outcome is a win
because he got better dice. The double doesn't have to occur at the latest
possible moment to clinch the game, therefore the game may or may not end with a
double.

I can't see that doubles would occur any more (or less) frequently on the last
roll than the first or somewhere in between.

I'm no mathematician and I am unable to 'prove it' one way or the other. All I
can do is think it through as best I can, but I must say that it does seems
counter-intuitive to say that more games end with a double. So I'm back to
'poppycock'.

What sayeth you?

--
JP White
Mailto:jp.w...@nashville.com

### Art Grater

Dec 13, 1999, 3:00:00 AM12/13/99
to
You don't need to be a statistician to visualize this....

Imagine a fight between a guy with a knife and a guy with boxing gloves.
Is the fight more likely to end with a gunshot or a punch?

### Art Grater

Dec 13, 1999, 3:00:00 AM12/13/99
to
Sorry, typo... I meant between a guy with a gun and a guy with boxing
gloves.

### Daniel Hollis

Dec 14, 1999, 3:00:00 AM12/14/99
to
In article <3855BDF8...@nashville.com>,
JP White <jp.w...@nashville.com> wrote:

>Chuck Bower wrote:
>
>My first thought when I read the proposition that more games end in doubles was
>'poppycock'.
>
>Then I started to think about it some more. It's not when the double comes along
>that's important, it's the effect it has. So I started to see the reasoning, a
>double will take more men off at the end of a game and may decide it. Hmmm
>
>Then I thought some more.
>
>Let's say two backgammon players had all their checkers in their respective home
>boards in exactly the same configuration. A non contact bearoff.
>
>Let's also assume both players roll identically (say consecutive 6-5's) and are
>both able to bearoff two men each roll.
>
>The first player on roll will win on his 8th roll.
>
>If on the 7th roll the second player on roll rolls 6-6 then he wins, before the
>first player can bearoff his last checker.
>
>So far so good, theory is holding up.
>
>But upon further analysis, it is clear that it doesn't matter *when* the 6-6 is
>rolled by the second player. 1st roll, 2nd roll, 3rd roll..the outcome is a win
>because he got better dice. The double doesn't have to occur at the latest
>possible moment to clinch the game, therefore the game may or may not end with a
>double.

The first player is going to get more chances to roll doubles again. Everyone's
always going to have a 1/6 chance to roll doubles - except for when the doubles
are rolled to bear off all your remaining men.

Consider a kind of 2-roll proposition in which both player A has 4 checkers on the
ace point, and player B has 2 checkers on the ace point.

If player A rolls doubles (1/6 time), game over
The next roll ends the game, always, and it's a double (10/36)(1/6) of the time.

The sum of these is (46/36)(1/6), which is greater than 1/6.

what the last roll of the game is, and roll doubles more often. This isn't so; the
dice will still roll doubles 1/6 of all rolls. What happens is that the last
roll isn't one specific roll: if the prop starts on the nth roll, then sometimes the
game ends on the nth roll, and sometimes it ends on the n+1 th. The probability of
doubles on the last roll is the probability of doubles on either of these rolls,
which bumps it up above 1/6.

An obvious extension of this is if I keep flipping coins until I get a heads.
The last roll will always be a heads, even though any particular flip
is only heads with probability 1/2. A similar thing is happening here.

Dan

### Stein Kulseth

Dec 14, 1999, 3:00:00 AM12/14/99
to

Chuck Bower wrote in message <833mc7\$cuj\$1...@flotsam.uits.indiana.edu>...

> In another thread ("Cooked Dice") there appeared to be a question
>about whether the last roll of a recorded game would contain bias, even
>if the dice generator itself were unbiased. E.g. is the probability that
>a game ends with doubles more than 1/6, less than 1/6, or exactly 1/6?

At least for games ending in races I think it is clear that it does.
If you get doubles, you are more likely to win the race, and if you get
doubles your last roll is more likely to be a double. Example, a pure
2 roll position will end with doubles 11/36 of the time.

For games ending with cube turns there might be a bias the other way, as
you usually get cubed out after a bad roll. Doubles are usually good rolls,
which might have allowed you to take the cube, or left your opponent not
strong
enough to double.

So there probably is a bias for or against doubles in the last roll, in fact
I
would be quite surprised if the two above effects cancelled out perfectly.
But it is not clear which way the bias is, and also it would matter whether
the
recorded game was money game or a game in a match - and then at what match
score.
For what it is worth, my guess is that the first effect dominates even at
Jacoby
money play and that doubles are slightly overrepresented in the last roll of
the game.

Note also that this does not mean that doubles are overrepresented overall.
Consider the following coinflipping game: We take turns flipping a coin, the
first to get "tails" win. Obviously tails are overrepresented in the last
flip
of the game, but this is countered by the fact that they are just as
seriously
underrepresented in all the other throws of each game. In total the
ratio should be 1:1 for a fair coin.

Similarily for the case of doubles in backgammon

> Consider the following proposition:
>
> A neutral third party will roll a pair of fair dice. Two players take
>on the following roles:
>
>If the dice roll comes up doublets, player A will pay player B five units.
>If the dice roll comes up NON-doublets, player B will pay player A one
unit.

Meaning this is a fair game, and both players have expectation 0 each roll

>Player A can call the game off at any time (after settling up, of course).
>Player B MUST continue to play the game as long as player A wants to.
>
> Assuming both players have sufficient bankrolls that neither is going
>to go broke, who has the advantage in this game?

I'd rather we didn't as such assumptions usually cause a lot of confusion,
but
OK, let's try.

Allowing B to have unlimited funds is unproblematic, he is the passive part
and will go along with whatever A does and collect or pay up in the end.

If we allow A to have unlimited funds he may choose to play until he is
allowing him to walk away with a small profit most of the time, but
sometimes being caught up playing for an indefinite time with an increasing
debt, leaving the game
ill-defined.
So we should assume that A's bankroll are limited, and to comply with the
assumption
that he isn't going to go broke, we must also assume that A puts an upper
bound on
his losses.

Then A has no expectation advantage in the game, the value to each player is
still 0.
However A may choose whether he wants to play for a big chance of small
profit vs a small chance of a big loss, or vice versa, or to play for equal
chance for equal profit/loss. This might make him feel good about it, but
the expectation is still
zero, which by the way beats the roulette.

> Would it matter if player

>B were the one who gets to decide when the game ends? What about if the

>neutral third party (dice roller) were the one who decided when the game
>ends?

No.

--
Stein Kulseth

### Kim Haubert

Dec 14, 1999, 3:00:00 AM12/14/99
to

Art Grater <in...@back-gammonNOSPAM.com> wrote in message
news:834gds\$t33\$1...@nntp6.atl.mindspring.net...

> You don't need to be a statistician to visualize this....
>
> Imagine a fight between a guy with a knife and a guy with boxing gloves.
> Is the fight more likely to end with a gunshot or a punch?
>
Must be the boxers bad day to get hit with a gun too.

### John Stryker

Dec 14, 1999, 3:00:00 AM12/14/99
to
On 13 Dec 1999 20:53:27 GMT, bo...@bigbang.astro.indiana.edu (Chuck
Bower) wrote:

> In another thread ("Cooked Dice") there appeared to be a question
>about whether the last roll of a recorded game would contain bias, even
>if the dice generator itself were unbiased. E.g. is the probability that
>a game ends with doubles more than 1/6, less than 1/6, or exactly 1/6?
>

> Consider the following proposition:
>
> A neutral third party will roll a pair of fair dice. Two players take
>on the following roles:
>
>If the dice roll comes up doublets, player A will pay player B five units.
>If the dice roll comes up NON-doublets, player B will pay player A one unit.

>Player A can call the game off at any time (after settling up, of course).
>Player B MUST continue to play the game as long as player A wants to.
>
> Assuming both players have sufficient bankrolls that neither is going

>to go broke, who has the advantage in this game? Would it matter if player

>B were the one who gets to decide when the game ends? What about if the
>neutral third party (dice roller) were the one who decided when the game
>ends?
>
>

> Chuck
> bo...@bigbang.astro.indiana.edu
> c_ray on FIBS

Just the thoughts of a layman...

I am not sure that this proposition touches on the material points of
the question, as the backgammon aspect has been removed. I think that
the last roll could be biased, but I don't know in which direction.
Consider recorded games in two groups:

1) Match is at double-match point (or one-point match); cubeless play

I think that the last roll would be biased in favor of doubles, as
they "move more freight" and are therefore more likely to move the
fifteenth man off.
Bias: doubles

2) Other matches and money play; cubeful play

I think that doubles are (all else being equal) more likely than
non-doubles to result in an equity shift that results in (double,
drop). Of these doubles, most would be the next-to-last roll rather
than the last roll (exception being dancing doubles), with a
non-double on the last roll to NOT recover the position.
Bias: non-doubles

If anyone has a sufficiently large database of recorded games (does
Hal Heinrich read this group?) I think that groups 1 and 2 would show
those biases.

Recorded games as a whole? I can't say. I don't have a feel for the
comparative strength of these biases against each other. I suspect
that the bias in group 1 is stronger, but that group 2 is a much
larger group, so I guess that the bias is against doubles. But this
is only a guess.
John Stryker
jo...@interaccess.com

### JP White

Dec 14, 1999, 3:00:00 AM12/14/99
to
Daniel Hollis wrote:

<snip>

> what the last roll of the game is, and roll doubles more often. This isn't so; the
> dice will still roll doubles 1/6 of all rolls. What happens is that the last
> roll isn't one specific roll: if the prop starts on the nth roll, then sometimes the
> game ends on the nth roll, and sometimes it ends on the n+1 th. The probability of
> doubles on the last roll is the probability of doubles on either of these rolls,
> which bumps it up above 1/6.

The mud is beginning to clear.

I wonder however, how often this situation occurs? The bias must be fairly small, not
that many games end in a neck and neck race.

The effect of the cube also confuses the issue since all games are not played to
completion, which probably reduces the effect of the bias some more. (Indeed you may be
less inclined to double if you opponent just rolled doubles which may be a bias the
other way). Hmmmm it's not as easy to figure out as it first appears. Though I agree
without the cube there is a bias.

Thanks for your explanation, it is the mathematical explanation I was unable to come up
with.

### Chuck Bower

Dec 14, 1999, 3:00:00 AM12/14/99
to
In article <833mc7\$cuj\$1...@flotsam.uits.indiana.edu>,
Chuck Bower <bo...@bigbang.astro.indiana.edu> wrote:

> In another thread ("Cooked Dice") there appeared to be a question
>about whether the last roll of a recorded game would contain bias, even
>if the dice generator itself were unbiased. E.g. is the probability that
>a game ends with doubles more than 1/6, less than 1/6, or exactly 1/6?
>
> Consider the following proposition:
>
> A neutral third party will roll a pair of fair dice. Two players take
>on the following roles:
>
>If the dice roll comes up doublets, player A will pay player B five units.
>If the dice roll comes up NON-doublets, player B will pay player A one unit.
>Player A can call the game off at any time (after settling up, of course).
>Player B MUST continue to play the game as long as player A wants to.
>
> Assuming both players have sufficient bankrolls that neither is going
>to go broke, who has the advantage in this game? Would it matter if player
>B were the one who gets to decide when the game ends? What about if the
>neutral third party (dice roller) were the one who decided when the game
>ends?

Hey, this is getting fun! Let's try some more.

Game # 2.

LV (or MC or...) casino sponsors a game. They (fairly) flip a perfect
coin and you get to bet A FIXED AMOUNT (let's say one local monetary unit)
on heads only. (That is, when you bet you are returned two monetary units
if it comes up heads and lose your monetary unit if it comes up tails.) If
you can play as long as you want (and both sides have sufficient bankrolls)
are you guaranteed to having the opportunity of walking away a winner?

Game # 3.

Same as game number 2, but house can "dial in" a small but finite
advantage (i.e. heads is slightly less likely than tails). How small must
they set the dial so that you are guaranteed to walk away a winner? (Or
can they?)

Game # 4.

(Let's get back to a more challenging game--backgammon!) You have
a truly random number generator-A create a sequence of rolls of a pair of
normal dice. You have a second truly random number generator-B create
non-doublet rolls. You have the two lists printed, and an impartial
3rd party (called 'ROLLER') goes SEQUENTIALLY through the lists, handing
out rolls in the following way:

i) At the beginning of the game, the NEXT roll from list B is chosen.
ii) From then until the end of the game, the NEXT roll from list A is used.
(Note that you play standard 'Western' backgammon with cubes, YOUR money :),
etc. It's just that instead of rolling your own dice you are using these

After a roll is used, it is crossed off and that line of the printed list
can never be used again. When a game ends, that roll is also HIGHLIGHTED
with a marking pen. When a new game begins, ROLLER must continue through
the lists sequentially.

If we just consider the A-list, then are the crossed off numbers
randomly distributed--in particular, are doubles likely to be 1/6 of
all the rolls? If the answer is 'no' (doubles should NOT occur with
an expectation of 1/6), what is their expectation? Before we used the
list we said it was fair. If doubles don't occur with an expectaion of
1/6, does that mean the list really isn't fair? Is it the last roll of
each game that is improper (that is, ALL the HIGHLIGHTED rolls) or only
the last roll crossed off the entire list? If between games, ROLLER
skipped an arbitrary number of lines=rolls (which aren't crossed of but
are still branded to never be used again), does that change your answer?
How about if ROLLER could skip an arbitrary number of lines between plays
during a game?

### Douglas Zare

Dec 14, 1999, 3:00:00 AM12/14/99
to
Chuck Bower wrote:

> In article <833mc7\$cuj\$1...@flotsam.uits.indiana.edu>,
> Chuck Bower <bo...@bigbang.astro.indiana.edu> wrote:
>
> > In another thread ("Cooked Dice") there appeared to be a question
> >about whether the last roll of a recorded game would contain bias, even
> >if the dice generator itself were unbiased. E.g. is the probability that
> >a game ends with doubles more than 1/6, less than 1/6, or exactly 1/6?
> >
> > Consider the following proposition:
> >
> > A neutral third party will roll a pair of fair dice. Two players take
> >on the following roles:
> >
> >If the dice roll comes up doublets, player A will pay player B five units.
> >If the dice roll comes up NON-doublets, player B will pay player A one unit.
> >Player A can call the game off at any time (after settling up, of course).
> >Player B MUST continue to play the game as long as player A wants to.
> >
> > Assuming both players have sufficient bankrolls that neither is going
> >to go broke, who has the advantage in this game? Would it matter if player
> >B were the one who gets to decide when the game ends? What about if the
> >neutral third party (dice roller) were the one who decided when the game
> >ends?

Each round is fair (0 expectation), so after any round, the expected advantage
of A over B is 0 regardless of the stopping strategy. It is possible to choose a
stopping strategy so that one stops only when A is losing. This would mean that,
if we are told that they are still playing, then on average A is winning. This
would be true by definition if they stop whenever B is winning, but we could
also stop only when B has won at least 100 units.

In real life, one would not be able to stop only when B is ahead, since the
average amoung of time one would have to wait is infinite, and the bankrolls are
really finite. There are plenty of "systems" people publish which amount to
ignoring this problem, and if you try to execute such a system in a casino it is
as though you are selling lottery tickets whose prize is your bankroll. It is
very likely that you gain a small amount, but eventually you lose your shirt;
from their perspective they see a few people go bankrupt on such systems each
day, and because of the house advantage this more than makes up for the majority
who win trivial amounts.

Thus, you can use casinos to give yourself risk in ways that are not immediately
obvious from the games, though the more complicated systems lose more because of
the house take. Some financial instruments have similar purposes, although there
the idea usually is to mitigate some risk.

> Hey, this is getting fun! Let's try some more.
>
> Game # 2.
>
> LV (or MC or...) casino sponsors a game. They (fairly) flip a perfect
> coin and you get to bet A FIXED AMOUNT (let's say one local monetary unit)
> on heads only. (That is, when you bet you are returned two monetary units
> if it comes up heads and lose your monetary unit if it comes up tails.) If
> you can play as long as you want (and both sides have sufficient bankrolls)
> are you guaranteed to having the opportunity of walking away a winner?

The exact same analysis holds.

> Game # 3.
>
> Same as game number 2, but house can "dial in" a small but finite
> advantage (i.e. heads is slightly less likely than tails). How small must
> they set the dial so that you are guaranteed to walk away a winner? (Or
> can they?)

Now, the expected amount of time before you are winning is finite. In practice,
you might still lose. The classical problem is to find the probability that you
are never losing. That is 2*prob_of_winning-1, e.g., if we bet even money on
whether doubles appear, the one who backs non-doubles has a 2/3 chance of never
being in debt. If the game is even or if you have a disadvantage, then with
probability 1 you will be losing at some point.

If you take a finite sample from A in any fashion which does not depend on
future rolls, then the average number of doubles will be 1/6. This is the case
if you stop after the second double, or take the first hundred but ignore the
rolls immediately after the doubles. If you sample from A in a way that depends
on future rolls, then the average might not be 1/6. This happens if you stop
just before the second double.

Even if the last roll in a backgammon game is more likely than 1/6 to be a
double, the non-initial rolls are still 1/6 doubles, and throwing out the last
roll will give an average lower than 1/6. That the last roll is more likely to
be a double seems like it should be the case just about any time it is not
cube-drop, e.g., no cube, cube-win, and cube-get-gammoned still seem like they
should end on a double more than 1/6 of the time. In the other situation, there
are a few times when doubles will provoke a cube-drop, such as when they save a
gammon, fail to come in against a potential blitz, or ruin a holding game, so I
think the overall rate of doubles on the last roll should be greater than 1/6 by
an observable amount.

Douglas Zare

### Steve Harris

Dec 14, 1999, 3:00:00 AM12/14/99
to
In article <833mc7\$cuj\$1...@flotsam.uits.indiana.edu>,
bo...@bigbang.astro.indiana.edu says...

> Consider the following proposition:
>
> A neutral third party will roll a pair of fair dice. Two players take
>on the following roles:
>
>If the dice roll comes up doublets, player A will pay player B five units.
>If the dice roll comes up NON-doublets, player B will pay player A one unit.
>Player A can call the game off at any time (after settling up, of course).
>Player B MUST continue to play the game as long as player A wants to.
>
> Assuming both players have sufficient bankrolls that neither is going
>to go broke, who has the advantage in this game? Would it matter if player
>B were the one who gets to decide when the game ends? What about if the
>neutral third party (dice roller) were the one who decided when the game
>ends?
>
>

> Chuck
> bo...@bigbang.astro.indiana.edu
> c_ray on FIBS

This would be a good test to give to an RNG (without the contest, of course).

Have it generate several rolls (some multiple of six), and keep score. After
several runs, the scores should be fairly close, and not biased either way (that
is, doublets and non-doublets should win roughly equal amounts of time).

--Steve

### Douglas Zare

Dec 15, 1999, 3:00:00 AM12/15/99
to
Douglas Zare wrote:

> Chuck Bower wrote:
>
> [...]

> > If we just consider the A-list, then are the crossed off numbers
> > randomly distributed--in particular, are doubles likely to be 1/6 of
> > all the rolls? If the answer is 'no' (doubles should NOT occur with
> > an expectation of 1/6), what is their expectation? Before we used the
> > list we said it was fair. If doubles don't occur with an expectaion of
> > 1/6, does that mean the list really isn't fair? Is it the last roll of
> > each game that is improper (that is, ALL the HIGHLIGHTED rolls) or only
> > the last roll crossed off the entire list? If between games, ROLLER
> > skipped an arbitrary number of lines=rolls (which aren't crossed of but
> > are still branded to never be used again), does that change your answer?
> > How about if ROLLER could skip an arbitrary number of lines between plays
> > during a game?
>
> If you take a finite sample from A in any fashion which does not depend on
> future rolls, then the average number of doubles will be 1/6. This is the case
> if you stop after the second double, or take the first hundred but ignore the
> rolls immediately after the doubles. If you sample from A in a way that depends
> on future rolls, then the average might not be 1/6. This happens if you stop
> just before the second double.

Gary Wong pointed out to me by e-mail that this is incorrect. Thanks. One correct
statement is that the expected number of heads will equal 1/6 of the expected
number of rolls. By stopping early we could cause small samples to be overweighted,
so the average of the values might not be 1/6.

Let me modify an example of his. Let's toss fair coins, and we will either stop
when there are more heads than tails in the first 3 tosses, or after 10^10 tosses.
If we stop early, which we do 1/2 of the time, the fraction of heads will be 1 or
2/3, and after 10^10 it will be very close to 1/2 with high probability. So we
would get an average over games of much greater than 1/2, but if we average over
tosses we will get 1/2.

If we compile the statistics by looking at a fixed number of rolls, then the
average should be 1/6. But it is more natural to consider some number of games, and
the average might not be 1/6. Interesting. As the number of games increases, I
believe the average will tend to 1/6.

Douglas Zare

### Mark Sproson

Dec 16, 1999, 3:00:00 AM12/16/99
to
Paul Tanenbaum wrote (in the "Cooked dice?" thread):

> Mark Sproson wrote:
> > What this means, though, is that if you didn't include the last roll
> > in your figures, you would get a lower-than-expected proportion of
> > doubles.
>
> In fact, the opposite may be the case, depending on how you define
> "last roll". See below.
>
> [snipped discussion about resignations in bear-off races]

Hmmm... interesting. To summarise, in this sort of situation, the loser
will probably resign BEFORE taking a non-double which puts the game out
of reach. On the other hand, s/he'll probably resign AFTER the opponent
rolling a double which does the same. So, in this situation, I'd guess
the last roll actually played is more likely to be a double than 1/6,
and the last roll thrown less likely.

But what proportion of games does this account for? I have no idea.
There must also be a fairly high proportion of games that end with a
dropped double - by definition, after the last roll has been played -
and it's harder to generalise about what the last roll would be.

How can we test this? Is there are large group of matches we could
analyse? They'd have to be non-biased - presumably, my collection of
games I saved because they were "interesting" is NOT a good sample!

> Now, our anonymous data statistician did indeed claim a larger-than-
> expected number of doubles. So I wonder: assume that when the trailer
> makes his final roll, and is mathematically eliminated, he resigns
> without playing it. Could it be the case that the GG files drop these
> rolls from their reports? Or could it be that His counting methodology
> behaved this way? That would explain part of the bias.

Could be. Aside from our friend's forthright arguments, the important
thing is that you MUST include all rolls made, whether they're taken or
not, to get analysable data. Some time back a study was carried out of
the odds of entering from the bar against a home board with n points
covered. A discrepancy was discovered - the observed incidence didn't
tally with the predicted. The discrepancy was later discovered to be
because of unplayed final rolls - once these were taken into account,
the numbers fitted.

> > Take an extreme case - on every roll, if it's a double it doesn't
> > count. Then count the proportion of doubles that are left - there'll
> > be 0 - a lot less than the 'predicted' 1/6 of all rolls!
> > OK, a middle case, every third roll, if it's a double it doesn't
> > count.
> > We'll lose one double every 18 rolls on average, so the incidence of
> > doubles will average at 2/17 - still rather lower than 1/6.
>
> This holds only because you are performing a 'conditional
> probability' experiment. You are 'peeking', so to speak, before
> deciding whether to omit the roll. This is significantly different
> than the action you propose below...

I perhaps didn't explain very clearly - I was just proposing short
games played in exactly the same way as normal BG. But I WAS wrong -
I didn't work through the maths of these samples properly.

Basically, by ignoring the first (and non-double) roll, you're also
ignoring any number of doubles that occurred before it. This amounts
to a sequence of rolls ending in a non-double, and the average incidence
of doubles within such sequences is the same as that in general - 1/6.

So from a sequence of numbers with a 1/6 proportion of doubles, we're
removing other sequences with the same proportion, leaving the overall
proportion unchanged.

Interesting to note, the average number of doubles "ignored" before
the start of a game is 1/5 - there's 1/6 chance of one double, 1/36
of a second, 1/216 of a third and so on. Summing this infinite series
gives you 1/5. Ignoring one (the first) non-double per game therefore
means you ignore five times as many non-doubles as doubles - exactly
the proportion we expect.

This may indeed be 'obvious' to some of you guys, but once I'd finally
worked it out I found it surprisingly 'neat'!

Sproz
Stats Victim Extraordinaire

### Chuck Bower

Dec 16, 1999, 3:00:00 AM12/16/99
to

My earlier 'games' were aimed at stimulating thought on backgammon
dice sampling. I'm now going to get more serious. There has been quite
a bit of discussion (and some of it intelligent :) on this topic in two
threads. I'm not going to assign names to ideas here, except maybe
some of Gary Wong's because he and I have been corresponding privately
and thus his ideas have not all been seen on this newsgroup.

The real question (at least the one I'm focussing on) is:

If one wants to look at backammon dice from real games and determine
if doubles occur with the proper expected frequency (defined by random
dice), how should the sample be taken?

One might think "it doesn't really matter", but that appears not to be
the case. For starters, Western backgammon rules state that the opening
roll is determined by each player tossing one die and the 'winner' playing
his/her die and the oppoenent's as the opening move of the game. Ties
here are rerolled. So the FIRST roll of a Western backgammon game should
be treated as a special case. If it were me, I would just eliminate it
from the sampling. If you know where the game started, this should be
fairly easy. I'm surprised that GamesGrid doesn't do this, but I'm sure
they have their reasons. I point out, though, that if you include the
first roll of the game, you CANNOT simply correct for it by some constant.
Correction factors would depend on the length of the game, and games have
different lengths.

OK, what about the last roll of each game? Does it matter if we
include it or not? Let's start off with a clearly defined case:

Assume you play Western BG, but with no cube, and no surrender of
any kind. The game ends when one player removes his/her last checkers
WITH A LEGAL DICE ROLL.

There has been a conjecture that a sample made up of these "last" dice
rolls has expectation of doubles greater than 1/6. Up to
now I don't believe anyone has provided a proof of this. In fact, if
you take a simpler subset--last dice rolls from games which end in
non-contact positions--even THAT has yet to be proved, I think. Gary
Wong and I have been discussing this (as have others, in posts to this
newsgroup). It certainly looks like the last diceroll in games ending
after contact has been broken has more than 1/6 probability of being
doubles, but the proof could be tough. Before challenging you to attempt
it, try the following:

Find a non-contact position where the probability of the game ending
in doubles is LESS THAN 1/6.

If such a position does not exist, then that might hint that a proof of
the above conjecture could be relatively straightforward. Unfortunately
Gary and I came up with a (simple) non-contact position where the probability
of the game ending with a double was 16.1%, which is less than 1/6. (I
could tell you the position now, but I think I'll wait and see if others
can figure it out, or even find a position with a SMALLER probability of
ending in doubles.) My conclusion is that the existence of such a
position just makes finding a proof to the conjecture difficult.

Even though the "last roll conjecture" hasn't been proved yet, let's
forge ahead. Let's ASSUME that the last roll of a cubeless, surrenderless
backgammon game has a probability > 1/6 of being a double. How does that
effect our sampling of backgammon dice? Should we throw out the last
double? Then are the remaining rolls randomly distributed?

I think here the answer is 'no'. In my previously posted "game 4"
where two (hypothetical) random number generators were used to makes lists
and then numbers were chosen from that list consecutively, I think I
hinted (but didn't prove) that if you include ALL dice rolls (other than
the opening rolls of every game), then these should be properly distributed.
Thus if you throw out the last roll of each game (which we are assuming
have P > 1/6 of being doubles) then the remaining dicerolls must have
on average P < 1/6 of being doubles. Gary is the one who first pointed
this out to me, and I think he's right.

Now, if all this is true, then by leaving off the last roll of the
game (even for cubeless, surrenderless Western backgammon) are you biasing
the remaining rolls? It appears the answer is 'yes'. As already hinted at
by Paul T., some (even maybe some robotic) recorders don't show the last
roll. For example, X has one checker on the 6-point and one on the 3-point
and the recorder says "X wins 2 points" but doesn't show the roll. If
this happens, then it seems that the sample is now biased. It also seems
on intitial contemplation that you cannot correct the sample to account
for this bias. You can probably assign bounds such that a true diceroll
bias could be detected if it goes in a certain direction, but that is getting
sticky.

So far we've only discussed cubeless, surrenderless Western BG.
But MOST matches include the cube (and specifically the double/pass
sequence) and some, as on FIBS, allow concessions. Do these
bias the sample? I certainly don't know for sure but it seems that

What's my conclusion? I may have more than one, but for sure
I think I can say that sampling backgammon dice in an unbiased fashion
is not such a simple task, and drawing conclusions from statistical
studies based on such sampling could be fraught with inaccuracies.
Note that the "dicetest" function on FIBS (and possibly analogous
commands on other servers) allow you to take an unbiased sample. Results
based on studies of these numbers won't satisfy BG conspiracy theorists
but right now I'm finding it hard enough to figure out a way of convincing
even rational observers. :)

### Gary Wong

Dec 16, 1999, 3:00:00 AM12/16/99
to
bo...@bigbang.astro.indiana.edu (Chuck Bower) writes:
> The real question (at least the one I'm focussing on) is:
>
> If one wants to look at backammon dice from real games and determine
> if doubles occur with the proper expected frequency (defined by random
> dice), how should the sample be taken?
>
> One might think "it doesn't really matter", but that appears not to be
> the case. For starters, Western backgammon rules state that the opening
> roll is determined by each player tossing one die and the 'winner' playing
> his/her die and the oppoenent's as the opening move of the game. Ties
> here are rerolled. So the FIRST roll of a Western backgammon game should
> be treated as a special case. If it were me, I would just eliminate it
> from the sampling. If you know where the game started, this should be
> fairly easy. I'm surprised that GamesGrid doesn't do this, but I'm sure
> they have their reasons.

Slightly better than ignoring the first roll (depending on your
purpose) would be to RECORD and reroll (rather than discard and
reroll) any doubles which come up as the opening roll.

> What's my conclusion? I may have more than one, but for sure
> I think I can say that sampling backgammon dice in an unbiased fashion
> is not such a simple task, and drawing conclusions from statistical
> studies based on such sampling could be fraught with inaccuracies.

I agree with everything you said, but even considering all the
pitfalls you point out, I still believe it is straightforward to make
unbiased samples as long as you're careful. The easiest way is just
to record EVERY SINGLE ROLL, regardless of whether it was rerolled to
start the game, whether it started or ended a game or was in between,
whether it was played or whether the player resigned after rolling it.
If you do that, you can be certain that your sample is expected to be
representative of the population; and your estimate of the probability
of doubles P(D) will be unbiased.

If you start discarding rolls from your sample (e.g. record every roll
but the last), you are no longer measuring P(D), but P(D|~E), the
probability that a roll was a double given that it did not end a game.
P(X|Y) does not equal P(X) in general; your estimate of P(X|Y) may
therefore be a biased estimator of P(X). Sometimes it will be
unbiased; P(X|Y) = P(X) if X and Y are independent. You can easily
show that P(D|~F), the probability that a roll is a double given that
it wasn't the first roll of the game, will equal P(D) as long as
you're willing to assume that D and F are independent. But on the
whole it's much easier to sample from the entire space, which relieves

Cheers,
Gary.
--
Gary Wong, Department of Computer Science, University of Arizona
ga...@cs.arizona.edu http://www.cs.arizona.edu/~gary/

### Daniel Hollis

Dec 17, 1999, 3:00:00 AM12/17/99
to
In article <83br8i\$6gg\$1...@flotsam.uits.indiana.edu>,

Chuck Bower <bo...@bigbang.astro.indiana.edu> wrote:
>
>If such a position does not exist, then that might hint that a proof of
>the above conjecture could be relatively straightforward. Unfortunately
>Gary and I came up with a (simple) non-contact position where the probability
>of the game ending with a double was 16.1%, which is less than 1/6. (I
>could tell you the position now, but I think I'll wait and see if others
>can figure it out, or even find a position with a SMALLER probability of
>ending in doubles.) My conclusion is that the existence of such a
>position just makes finding a proof to the conjecture difficult.

I think I see one: X has one checker left on his 2 point. O holds
X's one point, and has enough time to keep it. X can bear off with
every roll except for 11. So, the game will end on doubles will 22
through 66, and repeat on 11.

> Even though the "last roll conjecture" hasn't been proved yet, let's
>forge ahead. Let's ASSUME that the last roll of a cubeless, surrenderless
>backgammon game has a probability > 1/6 of being a double. How does that
>effect our sampling of backgammon dice? Should we throw out the last
>double? Then are the remaining rolls randomly distributed?

As Gary is pointing out with conditional probability, it won't really
matter if the last roll has a probability greater than or less than 1/6
of being a double. What matters is that the probability of a roll
being a double depends on whether or not that roll is the last
roll of the game; ie the two events are dependent.

Will the last-roll situations be such that the total probability of
getting a double is still 1/6? That's the same as saying that the
positions where it's >1/6 will exactly cancel out the positions where
it's <1/6, and though we can't really prove that, it seems ridiculous
to assume that it would.

The best answer, IMO, is to ask why we should bother figuring out
what the bias should be, when we leave out the last roll of the game,
when a easy alternative presents itself: just record all the rolls of
the game, including the last. (Of course, discussion on the
distribution of the last roll may be interesting on its own)

> I think here the answer is 'no'. In my previously posted "game 4"
>where two (hypothetical) random number generators were used to makes lists
>and then numbers were chosen from that list consecutively, I think I
>hinted (but didn't prove) that if you include ALL dice rolls (other than
>the opening rolls of every game), then these should be properly distributed.
>Thus if you throw out the last roll of each game (which we are assuming
>have P > 1/6 of being doubles) then the remaining dicerolls must have
>on average P < 1/6 of being doubles. Gary is the one who first pointed
>this out to me, and I think he's right.

This example I think demonstrated what's going on. If we highlight
the last roll of the game, we're not highlighting random rolls. If I
non-randomly highlighted all the rolls that included a 6, then I can't
assume that the distribution of doubles in the highlighted sample is
1/6. Conditional probability is necessary, and it will tell us that in
the highlighted sample, the probability of doubles is 1/11. The same
idea applies to highlighting the last rolls.

Dan

### dmg

Dec 17, 1999, 3:00:00 AM12/17/99
to
On 17 Dec 1999 17:58:59 GMT, hol...@math.umn.edu (Daniel Hollis)
wrote:

| This example I think demonstrated what's going on. If we highlight
|the last roll of the game, we're not highlighting random rolls. If I
|non-randomly highlighted all the rolls that included a 6, then I can't
|assume that the distribution of doubles in the highlighted sample is
|1/6. Conditional probability is necessary, and it will tell us that in
|the highlighted sample, the probability of doubles is 1/11. The same
|idea applies to highlighting the last rolls.
|
|Dan

I don't know anything about statistics and I'm barely able to follow
the discussion, so I probably shouldn't be jumping in here, but your
comments above made me wonder if this last roll bias (if it indeed
exists) would show itself in other ways as well. If the concept is
that doubles are more likely to end the game because they bear off
more checker combinations, wouldn't the same hold true (to a lesser
extent) for any large roll? You mentioned rolls that include a 6; if
the bias toward doubles exists, wouldn't there be a similar bias
toward rolls that include a 6, since they bear off checkers on lower
points (as opposed to smaller numbers which will often require another
throw of the dice)?
_____
dmg

### Chuck Bower

Dec 17, 1999, 3:00:00 AM12/17/99
to
In article <83dtl3\$odr\$1...@news1.tc.umn.edu>,
Daniel Hollis <hol...@math.umn.edu> wrote:

>In article <83br8i\$6gg\$1...@flotsam.uits.indiana.edu>,
>Chuck Bower <bo...@bigbang.astro.indiana.edu> wrote:
>>

>>If such a position does not exist, then that might hint that a proof of
>>the above conjecture could be relatively straightforward. Unfortunately
>>Gary and I came up with a (simple) non-contact position where the probability
>>of the game ending with a double was 16.1%, which is less than 1/6. (I
>>could tell you the position now, but I think I'll wait and see if others
>>can figure it out, or even find a position with a SMALLER probability of
>>ending in doubles.) My conclusion is that the existence of such a
>>position just makes finding a proof to the conjecture difficult.
>

> I think I see one: X has one checker left on his 2 point. O holds
>X's one point, and has enough time to keep it. X can bear off with
>every roll except for 11. So, the game will end on doubles will 22
>through 66, and repeat on 11.

Gary also found this one, although I think you have to include
what happens when O hits after X's 11. (When you do that you still find
that more than 5/6 of all games end in non-doublets, so your example
is a good one.) Actually I was trying to ask for a non-contact
position. I'll wait another day to give (the few who even care...) people
a chance to find one.

>> Even though the "last roll conjecture" hasn't been proved yet, let's
>>forge ahead. Let's ASSUME that the last roll of a cubeless, surrenderless
>>backgammon game has a probability > 1/6 of being a double. How does that
>>effect our sampling of backgammon dice? Should we throw out the last
>>double? Then are the remaining rolls randomly distributed?
>

> As Gary is pointing out with conditional probability, it won't really
>matter if the last roll has a probability greater than or less than 1/6
>of being a double. What matters is that the probability of a roll
>being a double depends on whether or not that roll is the last
>roll of the game; ie the two events are dependent.
>
> Will the last-roll situations be such that the total probability of
>getting a double is still 1/6? That's the same as saying that the
>positions where it's >1/6 will exactly cancel out the positions where
>it's <1/6, and though we can't really prove that, it seems ridiculous
>to assume that it would.
>
> The best answer, IMO, is to ask why we should bother figuring out
>what the bias should be, when we leave out the last roll of the game,
>when a easy alternative presents itself: just record all the rolls of
>the game, including the last. (Of course, discussion on the
>distribution of the last roll may be interesting on its own)

That is also what Gary said. But that may not be so easy. Yes,
if you write your own match recorder (for server-A or commercial robot-B).
But what I'm talking about is using a pre-existing match recorder (or even
pre-existing recorded matches) and analyzing them.

Two and a half years ago I asked Hal if it would be possible for him to
give me access so I could analyze HUMAN ROLLED DICE. He said 'yes' but
then warned that not all rolls were recorded. Specifically he said 10%
of matches didn't show dancing rolls, and some also didn't include other
non-playable rolls. At that time he pointed out that these data are
thus biased.

>> I think here the answer is 'no'. In my previously posted "game 4"
>>where two (hypothetical) random number generators were used to makes lists
>>and then numbers were chosen from that list consecutively, I think I
>>hinted (but didn't prove) that if you include ALL dice rolls (other than
>>the opening rolls of every game), then these should be properly distributed.
>>Thus if you throw out the last roll of each game (which we are assuming
>>have P > 1/6 of being doubles) then the remaining dicerolls must have
>>on average P < 1/6 of being doubles. Gary is the one who first pointed
>>this out to me, and I think he's right.
>

> This example I think demonstrated what's going on. If we highlight
>the last roll of the game, we're not highlighting random rolls. If I
>non-randomly highlighted all the rolls that included a 6, then I can't
>assume that the distribution of doubles in the highlighted sample is
>1/6. Conditional probability is necessary, and it will tell us that in
>the highlighted sample, the probability of doubles is 1/11. The same
>idea applies to highlighting the last rolls.

Since writing that post I realized that in some cases it may be
possible to CORRECT for the missing last rolls. Take the example I
gave: X has one checker on the 6-point, one on the 3-point, and the
recorder says: "X wins 2 points." You do have SOME information on this
roll, since you know it was sufficient to bear off the checkers. So
65, 64, 63 are the non-doublets (six total rolls) and 33, 44, 55, 66
are the doublets (four total rolls) which accomplish this. If you're
looking for doublets, then assign this last roll a value 0.4 doublets and
0.6 non-doublets. I think in this case the bias is removed. Still, to
actually program logic into some automated roll analyzer would likely
be a lot of effort (and probably not worth it).

I also thought of another way to illustrate how unbiased rolls can
result in games being more likely to end in doublets AND the consequence
of throwing out the last roll. Take the position which Dan used as an
example a few days ago:

123456
X-----| |
X | |
X | |
X |b|
|a|
|r|
| |
O | |
O-----| |
123456

X on roll.

Let's "play" this position out 1296 times with sequential dice (which
are not-so-pseudo-random :)

In the first 36 games, X rolls 11.
In the next 36 games, X rolls 21.
In the next 36 games, X rolls 12.
In the next 36 games, X rolls 31.
etc.

So, of these 1296 games, when X rolls doubles (which happens in 6 X 36 =
216 of the games), X wins on that roll and no more dice are rolled. In
the remaining 1080 games, O rolls as follows:

In the first 30 games O rolls 11.
In the next 30 games O rolls 21.
In the next 30 games O rolls 12.
In the next 30 games O rolls 31.
etc.

Of these 1080 remaining games, O rolls doubles 30 X 6 = 180 times. In
these 900 games O rolls singlets. In either case, all these games end
after two rolls.

Now, let's tally things up.

Total number of games: 1296
Total number of games ending in doubles: 216 + 180 = 396
Fraction of games which end in doubles: 396/1296 = 11/36 > 1/6

Total number of rolls: 1296 + 1080 = 2376
Total number of doubles: 216 + 180 = 396
Fraction of rolls which are doubles: 396/2376 = 1/6 (= 1/6 !)

So, our "random number generator" caused 30+% of games to end with doublets
but we verified that 1/6 of all rolls were doublets, as they were supposed
to be. NOW, what happens if we throw out the last roll? Are the remaining
rolls random (or more correctly, do they show the content of doublets and
non-doublets expected of random dice)? Well, the only rolls we didn't throw
out were the first rolls from the 1080 games which continued to a second
roll. NONE OF THESE ROLLS WAS a doublet. So the fraction of doublets in
this abbreviated sample is 0/1080 = 0, which is a far cry from the 1/6. :)

In summary, in this contrived example, if you look at all dice rolls
then the distribution of doublets and non-doublets was as expected. But
the distribution of doublets and non-doublets for only the last rolls of
the games contained a strong content of doubles (11/36). Consequently,
the non-last rolls of these games were under-represented in doublet content.
If you selectively throw out the last rolls, you bias an initially unbiased
sample.

Clearly this is a special case since not all backgammon games come
down to four checkers on the ace-point vs. two checkers on the acepoint.
I definitely have not proved that throwing out the last roll for typical
backgammon games will necessarily bias the dice. Hopefully I've given
evidence that it could be the case, so if it's OK to leave them in and
it might be wrong to throw them out, leave them in!

### Walter Trice

Dec 17, 1999, 3:00:00 AM12/17/99
to

Chuck Bower wrote in message <83br8i\$6gg\$1...@flotsam.uits.indiana.edu>...

[snip]

> Unfortunately
>Gary and I came up with a (simple) non-contact position where the
probability
>of the game ending with a double was 16.1%, which is less than 1/6. (I
>could tell you the position now, but I think I'll wait and see if others
>can figure it out, or even find a position with a SMALLER probability of
>ending in doubles.)

Hmm. If we give one player exactly one checker on his 5 point, then 5 rolls
don't bear off immediately, and 20% of them (11, from the set 11, 12, 21,
13, 31) are doubles. So let's give both players 1 man on the 5. Then the
chance that the game ends with doubles is:

5/36 + 5/36 * 5/36 + 1/6 * 5/36 * 5/36 = 16.14%

That the one?

[Note: the reason you have all those 5s in there is that, purely by
coincidence, 5 doublets immediately end the game, and another 5 rolls don't
bear off.]

-- walter trice

### Walter Trice

Dec 18, 1999, 3:00:00 AM12/18/99
to
The probability that the roll that ends the game is doubles can be precisely
calculated for simple models of the race. Consider the ace-point stack.
Assume that you start with some stack of checkers sufficiently huge that the
probabilities of hitting any particular n-roll position, for "small" values
of n, may be considered equal. Let e(n) denote the probability that in a
particular bearoff an n-roll position is ever reached.

Now, the only way not to reach an n-roll position is to reach an (n+1)-roll
position and then roll doubles. Hence e(n) = 1 - e(n+1)/6. But we assume all
values of e(n) equal, so e(n) = 1 - e(n)/6. Solving this, we get e(n) = 6/7.

The ace-point-stack bearoff will be completed by rolling doubles if doubles
are rolled from either a 2-roll position or a 1-roll position. The
probability that this happens is 2*6/7*1/6, or about 28.6%.

That may seem large, but just to check I did a little manual rollout,
repeatedly bearing off a stack of 15 checkers. The roll that bore off the
last checker (sometimes the last 3 checkers) was a doublet 31% of the time
in 108 trials. Seems close enough :-)

You could get a similar estimate for the 1-checker model. The average pips
per roll is 8&1/6, or 49/6, and this turns out to imply that the chance of
achieving an n-pip position is 6/49 for "small" n, given that the pipcount
you started with was "large." So you could calculate the probability that
doubles ends a one-sided race by adding up the numbers for pipcounts from 1
through 24. I.e. you would have 6/49 * [1/36 * 4 + 2/36 * 4 + 3/36 * 4 +
4/36 * 4 + 5/36 * 4 + 6/36 * 4] = 6/49 * 1/36 * 4 * 21 = 28.6%. Well,
whaddayaknow, same as for the ace-point stack. (Since I didn't know this
before just now, I can't explain it! Somebody please put this into a
generalized model for me so I don't have to do some actual thinking...)

The experiment of repeatedly rolling out the 15-checker ace-point-stack, I
think, sheds a bit of light on the whole question of whether this last-roll
thing results in any bias when dice-rolls that result from BG games are
observed. Suppose you roll out the one-sided 15-checker bearoff repeatedly.
In the long run, you will find that game-ending rolls are doubles 28.6% of
the time. But it should also be obvious that over the long run doubles will
occur 1/6 of the time. The dice don't "know" that checkers are being
shuffled as a result of how they come down!

To clarify the thought-experiment (if necessary) you could imagine one guy
in a room alone rolling dice, calling out the numbers, and recording them,
while the checkers are being moved in another room. The person moving the
checkers can hear what the rolls are, but the roller has no knowledge of
which rolls end bearoffs. From the point of view of the roller, he is just
rolling dice interminably, and of course he is going to roll doubles about
1/6 of the time.

What this seems to tell us is that, yes, the last roll in a game of
backgammon is going to be doubles more than 1/6 of the time, but, no, this
does not result in any over-all bias towards doubles when you observe
dice-sequences generated by backgammon games. The reason will turn out to be
that rolls that fail to end games will be doubles less often than 1/6 of the
time.

Question for further analysis: by any chance is it true that in real
backgammon the game-ending roll is a doublet with probability .2857142857? I
suspect that the answer is not exactly, but pretty close, and perhaps very
close. (Of course we are just talking about cubeless play.)

-- walter trice

Note: when I refer to an n-roll position, I mean a position consisting of
either 2n or 2n-1 checkers on an acepoint.

### Douglas Zare

Dec 18, 1999, 3:00:00 AM12/18/99
to
Walter Trice wrote:

> The probability that the roll that ends the game is doubles can be precisely

> calculated for simple models of the race. [...]

>
> You could get a similar estimate for the 1-checker model. The average pips
> per roll is 8&1/6, or 49/6, and this turns out to imply that the chance of
> achieving an n-pip position is 6/49 for "small" n, given that the pipcount
> you started with was "large." So you could calculate the probability that
> doubles ends a one-sided race by adding up the numbers for pipcounts from 1
> through 24. I.e. you would have 6/49 * [1/36 * 4 + 2/36 * 4 + 3/36 * 4 +
> 4/36 * 4 + 5/36 * 4 + 6/36 * 4] = 6/49 * 1/36 * 4 * 21 = 28.6%. Well,
> whaddayaknow, same as for the ace-point stack. (Since I didn't know this
> before just now, I can't explain it! Somebody please put this into a

> generalized model for me so I don't have to do some actual thinking...)[...]

Nice arguments. I was surprised by this, but I found the following helpful: If
you roll all 36 combinations, you get to play 84 numbers: 10 of each number in
non-doubles, and 4 of each number in doubles. So 2/7 of the numbers you get to
play are from doubles. Similarly, 2/7 of the pipcount is from doubles. Instead
of taking a random starting point and a fixed finish line, you might think of a
fixed starting point and a random finish line. Whether you choose the finish
line by a certain number of checkers removed or by a certain point being
passed, 2/7 of the action comes from doubles, so the chance that the finish
line is within a double is 2/7.

On the other hand, this is not a one-sided race. A side rolling non-doubles
will tend not to be the side rolling last. This effect shinks as the length of
the race increases, but consider a 2-roll vs. 2-roll position. The only way
that this will not end in doubles is for there to be 3 consecutive non-doubles,
(5/6)^3=125/216. So this will end in doubles not 2/7 of the time but 91/216
~42.1% of the time.

Douglas Zare

### Mark Sproson

Dec 18, 1999, 3:00:00 AM12/18/99
to

Douglas Zare wrote:

> Nice arguments. I was surprised by this, but I found the following helpful: If
> you roll all 36 combinations, you get to play 84 numbers: 10 of each number in
> non-doubles, and 4 of each number in doubles. So 2/7 of the numbers you get to
> play are from doubles. Similarly, 2/7 of the pipcount is from doubles. Instead
> of taking a random starting point and a fixed finish line, you might think of a
> fixed starting point and a random finish line. Whether you choose the finish
> line by a certain number of checkers removed or by a certain point being
> passed, 2/7 of the action comes from doubles, so the chance that the finish
> line is within a double is 2/7.

I had a look at positions where player on roll has two remaining men -
for all possible positions of the two men - and the opponent has one
man on the one-point (or equivalent). Here's the chances of the game
ending on a double, for each position. (I've assumed the best play
if there's any choice).

1,1 0.167
1,2 0.167
1,3 0.176
1,4 0.171
1,5 0.199
1,6 0.236
2,2 0.213
2,3 0.190
2,4 0.199
2,5 0.218
2,6 0.245
3,3 0.227
3,4 0.227
3,5 0.213
3,6 0.231
4,4 0.255
4,5 0.231
4,6 0.241
5,5 0.250
5,6 0.250
6,6 0.259

If player on roll has only 1 man the chances for each position are

1 0.167
2 0.167
3 0.167
4 0.176
5 0.162
6 0.181

I also looked at what happened when both players had two men on 6,6.
There's only a 4.5% chance that this position will last to a fifth roll,
and a 25.3% chance that the game will end in a double before then - I'm
surprised to see that this figure is lower than the value for 6,6 in the
table above.

Similar calculations with both players on 4,4 yield a 26.5% chance of
the game ending in a double before the fifth roll with only a 0.04%
chance of reaching a fifth roll.

I was going to conjecture an upper bound for any (non-resignation)
position of around 27%, then saw Walter Trice's figure of 28.6% - so
I'd say that what I've worked through supports Walter's conjecture.
Except...

> On the other hand, this is not a one-sided race. A side rolling non-doubles
> will tend not to be the side rolling last. This effect shinks as the length of
> the race increases, but consider a 2-roll vs. 2-roll position. The only way
> that this will not end in doubles is for there to be 3 consecutive non-doubles,
> (5/6)^3=125/216. So this will end in doubles not 2/7 of the time but 91/216
> ~42.1% of the time.

Damn! There goes my "upper bound"! I have to say, positions where ALL
doubles and NO non-doubles finish the game are pretty rare (there's
only three of them!) - typical for 3-man positions is the case where
all doubles except 1-1 finish, giving games ending in doubles about 38%
of the time.

Sproz

### Chuck Bower

Dec 18, 1999, 3:00:00 AM12/18/99
to
In article <FMwqn...@world.std.com>, Walter Trice <w...@world.std.com> wrote:
>
>Chuck Bower wrote in message <83br8i\$6gg\$1...@flotsam.uits.indiana.edu>...
>
>
> [snip]
>
>> Unfortunately
>>Gary and I came up with a (simple) non-contact position where the
>probability
>>of the game ending with a double was 16.1%, which is less than 1/6. (I
>>could tell you the position now, but I think I'll wait and see if others
>>can figure it out, or even find a position with a SMALLER probability of
>>ending in doubles.)
>
>Hmm. If we give one player exactly one checker on his 5 point, then 5 rolls
>don't bear off immediately, and 20% of them (11, from the set 11, 12, 21,
>13, 31) are doubles. So let's give both players 1 man on the 5. Then the
>chance that the game ends with doubles is:
>
> 5/36 + 5/36 * 5/36 + 1/6 * 5/36 * 5/36 = 16.14%
>
>That the one?
>
>[Note: the reason you have all those 5s in there is that, purely by
>coincidence, 5 doublets immediately end the game, and another 5 rolls don't
>bear off.]

That's the one we found. Can anyone do better? If not, can anyone
PROVE that this is the extreme case? (Not that it's worth the effort...)

The "one on the five vs. one on the five" counterexample comes about
because 11 is such a wimpy roll (and the only wimpy doublet). If you just
consider one checker (don't worry about the opponent) and start by putting
it on the 24-point and counting the rolls which bear it off aTHIS URN and
what the doubles content of those successful rolls is, I think you find that
the ONLY place you can put that checker where doublets are UNDER-represented
is on the 5-point.

Still, I think I'll keep 11 in my arsenal. It can be a pretty nice
roll in contact positions. :)

### Gary Wong

Dec 18, 1999, 3:00:00 AM12/18/99
to
bo...@bigbang.astro.indiana.edu (Chuck Bower) writes:
> Walter Trice <w...@world.std.com> wrote:
> >Chuck Bower wrote in message <83br8i\$6gg\$1...@flotsam.uits.indiana.edu>...
> >> Unfortunately
> >>Gary and I came up with a (simple) non-contact position...
> >
> >Hmm. If we give one player exactly one checker on his 5 point...
> >
> >That the one?

>
> That's the one we found.

Chuck is being too kind here. He found it; I only talked about it.

He's also being discreet about the fact that he found it immediately
after I said I didn't think it could be done ;-)

### Joern Thyssen

Dec 20, 1999, 3:00:00 AM12/20/99
to

I let GNU Backgammon play against itself 13483 games and recorded all
dice rolls.
29.8% of all games ended with doubles.
^^^^^

Here is the recorded data if you want to crunch the numbers yourself or
calculate appropriate statistical measures.

GNU backgammon playing on 0-ply (equiv. to Snowie 1-ply). All rolls
counted, including initial doubles. No doubling cube used. The dice
generator was a pseudo-random one - if someone insists I can repeat the
experiment using true random numbers from <URI:"http://www.random.org">.

(gnubg) show dicestat
Dice statistics:

Die Number Freq
1 227434 0.16702
2 227405 0.16700
3 225959 0.16593
4 226852 0.16659
5 227094 0.16677
6 227000 0.16670
---------------------------------
Total 1361744 1.00000

Roll statistics...

Roll Number Freq
11 18905 0.02777
21 38007 0.05582
22 19099 0.02805
31 37650 0.05530
32 37774 0.05548
33 18635 0.02737
41 38084 0.05593
42 37857 0.05560
43 37790 0.05550
44 18766 0.02756
51 37945 0.05573
52 37536 0.05513
53 37692 0.05536
54 38074 0.05592
55 18846 0.02768
61 37938 0.05572
62 38033 0.05586
63 37783 0.05549
64 37515 0.05510
65 38155 0.05604
66 18788 0.02759
---------------------------------
Total 680872 1.00000

Doubles...

Total number of rolls : 680872
Number of doubles: 113039 [16.602%]

Games ending with doubles...

Total number of games : 13483
Games ending with doubles: 4013 [29.763%]

Joern

--
Joern Thyssen <j.thyssen(at)auckland.ac.nz>

### Walter Trice

Dec 20, 1999, 3:00:00 AM12/20/99
to
Very interesting! This supports my 2/7 estimate + the upward bias predicted
by Douglas Zare. Thanks!

-- walter trice

Joern Thyssen wrote in message
<385D6C25.325BC87E@_nospam_.auckland.ac.nz>...

>
>I let GNU Backgammon play against itself 13483 games and recorded all
>dice rolls.
>29.8% of all games ended with doubles.
>^^^^^
>

[major snip]

### Daniel Hollis

Dec 20, 1999, 3:00:00 AM12/20/99
to
In article <83eecb\$edg\$1...@flotsam.uits.indiana.edu>,

Chuck Bower <bo...@bigbang.astro.indiana.edu> wrote:
>> The best answer, IMO, is to ask why we should bother figuring out
>>what the bias should be, when we leave out the last roll of the game,
>>when a easy alternative presents itself: just record all the rolls of
>>the game, including the last. (Of course, discussion on the
>>distribution of the last roll may be interesting on its own)
>
> That is also what Gary said. But that may not be so easy. Yes,
>if you write your own match recorder (for server-A or commercial robot-B).
>But what I'm talking about is using a pre-existing match recorder (or even
>pre-existing recorded matches) and analyzing them.

My intuition is that this isn't really going to be possible with our
current understanding of the game, but you might be able to get an idea
with a neural net simulation.

> Clearly this is a special case since not all backgammon games come
>down to four checkers on the ace-point vs. two checkers on the acepoint.
>I definitely have not proved that throwing out the last roll for typical
>backgammon games will necessarily bias the dice. Hopefully I've given
>evidence that it could be the case, so if it's OK to leave them in and
>it might be wrong to throw them out, leave them in!

We can show that the tail end of a game can have this last-roll bias.
To show that not including the last roll of /all/ games induces a bias
reduces to showing how often a general game comes down to this position
versus all other positions with a last-roll bias, so that we can average
the total bias.
What I'm using here is the conditional probability formula of
P(E) = P(E|F) + P(E|CF). We've computed P(last roll doubles, given a 2
roll proposition), but to get the general bias for doubles on the last
roll, we need to compute the bias for all other positions.

As I was saying, this isn't going to be possible with our understanding
of the game. I think we'll have to wait until computers have done the
deterministic computation of all the positions of the game, in some sort
of markov-chain type computation. That's got to take virtually forever,
unfortunately.
And, I don't think there's an easier way to get the true numbers. You
might be able to make do with a neural net bot, and just record the prob.
of last-roll doubles over a long number of games. However, I am guessing
that the bias is going to be a small number; it might be caught in the
noise of the simulation. Even worse, a particular bot's style might
affec the results. A bot more inclined to blitz is going to be less
likely to race, and I think this bias will be greater in race positions.
Of course, I'm just guessing here.

Dan

### Daniel Hollis

Dec 20, 1999, 3:00:00 AM12/20/99
to
In article <385D6C25.325BC87E@_nospam_.auckland.ac.nz>,

Joern Thyssen <joern@_nospam_.thyssen.nu> wrote:
>
>I let GNU Backgammon play against itself 13483 games and recorded all
>dice rolls.
>29.8% of all games ended with doubles.
>^^^^^
>

This is larger than I expected.

It also lends itself to the race vs. blitz issue that I just brought up
in response to Mr. Bower: GNU only plays 1 point matches for the time
being. I am fairly certain that if you play it in a 1 point match, it
knows not to take extra risks for a gammon. In 1 point matches, blitz
play is a less attractive strategy than in money play, and so a race is
more likely.

Basically, relating this number to general games isn't going to be so
easy, but that's not so bad - 1 point matches have their own special
last-roll bias anyway. Since we can't say that X% of games are 1
pointers, we'd have to break down the bias that we're trying to measure
into biases for different scores.

Dan

### Gary Wong

Dec 20, 1999, 3:00:00 AM12/20/99
to
Joern Thyssen <joern@_nospam_.auckland.ac.nz> writes:

> Daniel Hollis wrote:
> > It also lends itself to the race vs. blitz issue that I just brought up
> > in response to Mr. Bower: GNU only plays 1 point matches for the time
> > being. I am fairly certain that if you play it in a 1 point match, it
> > knows not to take extra risks for a gammon. In 1 point matches, blitz
> > play is a less attractive strategy than in money play, and so a race is
> > more likely.
>
> Well, the relevant piece of code is:

<snip>

> That is, gnubg does actually play for gammon and backgammon.
>
> The version of gnubg playing on FIBS under the name 'gnu' probably (Gary
> Wong would have to clarify this) use
>
> extern float Utility( float ar[ NUM_OUTPUTS ] ) {
>
> return ar[ OUTPUT_WIN ] * 2.0 - 1.0;
> }

Right (well, actually it sets the gammon price to 0, which has
precisely the same effect). "mgnutest" is a different version of gnu
by Joseph Heled which plays longer matches on FIBS and sets the gammon
price according to the match score.

> But I also believe that the actual percentage of games ending in doubles
> could be very different from player to player. Other mails in this
> thread treats bearoff, but what about backgames, ace point holding games
> etc? Each type of game would have different percentage of games ending
> in doubles.

True, but I don't they would change TOO much. Even the exotic
examples we have come up with only vary by around +/-10%; I don't
think that the averages of classes overall would vary by more than 5%
(e.g. the probability of a back game position ending in doubles
matches the probability of a race ending in doubles by +/-5%). Either
a shot is hit (in which case we probably see a containment position
and then a race), or no shot is hit, in which case the game ends by
bearing off. And in this case (as in all bearoffs), the game is
generally more likely to end in doubles because doubles tend to take
more men off, whether against contact or not. And players behave
similarly enough that I would expect any reasonable player (i.e. one
that's not playing suicidally or otherwise unusually) to end in
doubles with a probability within (say) 2% of the typical player
(which we approximate with gnubg).

### Joern Thyssen

Dec 21, 1999, 3:00:00 AM12/21/99
to
Daniel Hollis wrote:
>
<snip>

>
> This is larger than I expected.
>
> It also lends itself to the race vs. blitz issue that I just brought up
> in response to Mr. Bower: GNU only plays 1 point matches for the time
> being. I am fairly certain that if you play it in a 1 point match, it
> knows not to take extra risks for a gammon. In 1 point matches, blitz
> play is a less attractive strategy than in money play, and so a race is
> more likely.
>
> Basically, relating this number to general games isn't going to be so
> easy, but that's not so bad - 1 point matches have their own special
> last-roll bias anyway. Since we can't say that X% of games are 1
> pointers, we'd have to break down the bias that we're trying to measure
> into biases for different scores.

Well, the relevant piece of code is:

extern float Utility( float ar[ NUM_OUTPUTS ] ) {

return ar[ OUTPUT_WIN ] * 2.0 - 1.0 +
ar[ OUTPUT_WINGAMMON ] * arGammonPrice[ 0 ] -
ar[ OUTPUT_LOSEGAMMON ] * arGammonPrice[ 1 ] +
ar[ OUTPUT_WINBACKGAMMON ] * arGammonPrice[ 2 ] -
ar[ OUTPUT_LOSEBACKGAMMON ] * arGammonPrice[ 3 ];
}

with arGammonPrice set appropriately ( arGammonPrice = {1.0, 1.0, 1.0,
1.0}).

That is, gnubg does actually play for gammon and backgammon. So my
experiment was a money game session played without cube - ie. all games
played to the very end. So gnubg would actually blitz if appropriate and
try to avoid being gammoned etc.

The version of gnubg playing on FIBS under the name 'gnu' probably (Gary
Wong would have to clarify this) use

extern float Utility( float ar[ NUM_OUTPUTS ] ) {

return ar[ OUTPUT_WIN ] * 2.0 - 1.0;

}

as the gammon price is zero for a 1-point match.

But I also believe that the actual percentage of games ending in doubles
could be very different from player to player. Other mails in this
etc? Each type of game would have different percentage of games ending
in doubles.

Joern

### David Smyth

Dec 21, 1999, 3:00:00 AM12/21/99
to
I just came into this thread and I can't be bothered reading it in its
entirety.

However I will assert that repeatedly running a random number
generator and stopping it on a specific number on each run will, of
course, *not* bias the final statistics toward a greater preponderance
of that particular number.

If you consider that the overall statistics (which is what we are
interested in) is simply all the individual runs concatenated together
very easily proves this. All we have effectively done is break up a
lengthy run of the random number generator on a particular number then
joined the segments back together again.

This is, of course, still true if we were to run the random number
generator between segments discarding an arbitrary number of random
numbers. A bias will only be produced if we specifically discard
numbers until a desired number is generated. This is clearly *not*
the case when forcing the run to end on a particular number as long as
we don't discard the numbers generated prior to the arrival of the
target number.

On 13 Dec 1999 20:53:27 GMT, bo...@bigbang.astro.indiana.edu (Chuck

Bower) wrote:

> In another thread ("Cooked Dice") there appeared to be a question
>about whether the last roll of a recorded game would contain bias, even
>if the dice generator itself were unbiased. E.g. is the probability that
>a game ends with doubles more than 1/6, less than 1/6, or exactly 1/6?
>

> Consider the following proposition:
>
> A neutral third party will roll a pair of fair dice. Two players take
>on the following roles:
>
>If the dice roll comes up doublets, player A will pay player B five units.
>If the dice roll comes up NON-doublets, player B will pay player A one unit.
>Player A can call the game off at any time (after settling up, of course).
>Player B MUST continue to play the game as long as player A wants to.
>
> Assuming both players have sufficient bankrolls that neither is going
>to go broke, who has the advantage in this game? Would it matter if player
>B were the one who gets to decide when the game ends? What about if the
>neutral third party (dice roller) were the one who decided when the game
>ends?
>
>

### Daniel Hollis

Dec 21, 1999, 3:00:00 AM12/21/99
to
In article <wt66xtp...@brigantine.CS.Arizona.EDU>,

Gary Wong <ga...@cs.arizona.edu> wrote:
> And in this case (as in all bearoffs), the game is
>generally more likely to end in doubles because doubles tend to take
>more men off, whether against contact or not. And players behave
>similarly enough that I would expect any reasonable player (i.e. one
>that's not playing suicidally or otherwise unusually) to end in
>doubles with a probability within (say) 2% of the typical player
>(which we approximate with gnubg).

If we have Bruce Becker playing, it's going to affect the results. :-).

We, of course, will have more doubles on the last roll due to the
previously discussed phenomenon. I'm just commenting that it's going to
be even more likely if both players have a shot at winning. For instance,
4 checkers on the ace point for each player gives the dice two chances to
roll doubles before the game must end. In order for the game not to end
on a doubles, non doubles must be rolled 3 times in a row - that's a
probability of (30/36)^3 = .5787 .

I think there are lots of chances, even if someone isn't slotting
wildly, for personal style to come in. Lots of games have a pay now vs
pay later decision where getting hit means the game. If a player is more
inclined to pay now, then the last-roll bias goes down compared to optimal
play.

Dan

### Chuck Bower

Dec 21, 1999, 3:00:00 AM12/21/99
to
In article <385f840e...@news.uq.edu.au>,
David Smyth <dav...@qimr.edu.au> wrote:

>I just came into this thread and I can't be bothered reading it in its
>entirety.

Well, EXCUUUUUUUUUUSE US!

>However I will assert that repeatedly running a random number
>generator and stopping it on a specific number on each run will, of
>course,

...a horse, of course! He'll give you the answer that you endorse...

>*not* bias the final statistics toward a greater preponderance
>of that particular number.
>
>If you consider that the overall statistics (which is what we are
>interested in) is simply all the individual runs concatenated together
>very easily

Yes, VERY easily...

>proves this. All we have effectively done is break up a
>lengthy run of the random number generator on a particular number then
>joined the segments back together again.
>
>This is, of course,

...he's always on a steady course...

>still true if we were to run the random number
>generator between segments discarding an arbitrary number of random
>numbers. A bias will only be produced if we specifically discard
>numbers until a desired number is generated. This is clearly *not*

Clearly *not*...

>the case when forcing the run to end on a particular number as long as
>we don't discard the numbers generated prior to the arrival of the
>target number.

So, let me get this straight. You "can't be bothered reading (this
thread) in its entirity", but you do have the time to write up an argument
which has been stated two or three times already in this thread. Are
we supposed to impressed with your time management technique?

Finally, if you "can't be bothered reading" what others have to say,
why should we be bothered reading what YOU have to say? Oh, I see,
because your time is more valuable than ours. Clearly, of course!

### Daniel Hollis

Dec 21, 1999, 3:00:00 AM12/21/99
to
In article <385f840e...@news.uq.edu.au>,
David Smyth <dav...@qimr.edu.au> wrote:
>
>However I will assert that repeatedly running a random number
>generator and stopping it on a specific number on each run will, of
>course, *not* bias the final statistics toward a greater preponderance
>of that particular number.

Yup. However, the issue that we're discussing is essentially stopping
the generator on an /nonspecific/ number, that is, after a specific
event - the last roll of the game.

As has been said before, and as a parallel situation, consider
stopping the generator after the nth roll that includes a 6. It's not a
specific roll, and the dice don't have the same distribution as they
usually do.

Dan

### JP White

Dec 21, 1999, 3:00:00 AM12/21/99
to
David Smyth wrote:

> I just came into this thread and I can't be bothered reading it in its
> entirety.
>

> However I will assert that repeatedly running a random number
> generator and stopping it on a specific number on each run will, of
> course, *not* bias the final statistics toward a greater preponderance
> of that particular number.
>

You DO need to read the thread. I, like you, was of the opinion that no bias
could exist for
essentially the reasons you have given. I have been convinced otherwise.

If we did not move 4 checkers when we threw a double then all you have said is
true enough and no bias would exist.
The act of moving 4 checkers for a double creates a game winning bias in last
throw and next to last throw scenarios..

<rest of post zapped (but I did read it lol)>

--
JP White
Mailto:jp.w...@nashville.com

### Walter Trice

Dec 22, 1999, 3:00:00 AM12/22/99
to

JP White wrote in message <3860207C...@nashville.com>...

But all he is saying is that there is no bias in the aggregate, which is
true.

Incidentally, there is a discussion of one of the themes of this thread in
Richard Epstein's book "Theory of Gambling and Statistical Logic." He gives
a bunch of theorems and corollaries, and one of the corollaries is stated as
follows: "No advantage accrues to the gambler who possesses the option of
discontinuing the game after each play." (Page 57 of the 2nd edition.) This
corollary directly implies that the dice rolls in a game of backgammon will
have the same expectations as for a single roll (i.e., no "bias" in the
aggregate.) The formal proof involves measure theory, infinite-dimensional
spaces, and that sort of thing. (Also, just in case anybody really wants to
try to track this down, the proof is not actually in the 2nd edition. A
footnote implies that the proofs are in the first edition (!)).

-- Walter Trice

### Douglas Zare

Dec 22, 1999, 3:00:00 AM12/22/99
to
Joern Thyssen wrote:

> [...]

>
> But I also believe that the actual percentage of games ending in doubles
> could be very different from player to player. Other mails in this
> thread treats bearoff, but what about backgames, ace point holding games
> etc? Each type of game would have different percentage of games ending
> in doubles.

Thanks for the data from the rollouts.

How well does GNU play Nackgammon, or a hybrid mixture? The 2/7 should be
robust, but the excess over 2/7 may vary. I would guess the variation would be
from a small fraction of a percent to ~3%. (It can be increased slightly by
irrational play, e.g., if I could win but decide not to bear both checkers
off.)

On the other hand, the two situations Walter Trice analyzed initially give
quite different answers to dmg's question about the bias toward 6's. If all
checkers are on the ace point there is no bias toward 6's, but with one
checker there will be a 6/(1+2+...+6)=2/7 bias again. The bias toward 6's
probably depends strongly on the style of play. It might be more helpful to
think of the bias against 1's and 2's rather than the bias toward 6's; I doubt
that the bias toward 6's over 5's is close to the bias toward 6's over 1's. On
the other hand, it should make quite a difference whether one tends to bear
off with contact or race.

Would you happen to have the distribution on the ending rolls of the 13483
games?

Douglas Zare

### DJWhitfill

Dec 23, 1999, 3:00:00 AM12/23/99
to
Is this another way of saying that the dice have no memory?

=============================
Subject: Re: sampling backgammon dice
From: "Walter Trice" w...@world.std.com
Date: Wed, 22 December 1999 01:00 PM EST
Message-id: <Fn5M3...@world.std.com>

### Walter Trice

Dec 23, 1999, 3:00:00 AM12/23/99
to
DJWhitfill wrote in message
<19991222224339...@ng-cr1.aol.com>...

>Is this another way of saying that the dice have no memory?

Not quite. It is more related to the idea that when the dice have no memory,
you can't change the odds by "knowing when to quit." There is a difference,
and even people who have a pretty good understanding of what statistical
independence means can fall into the fallacy. Here's an example:

"Operation of a gambling house or hell is a notoriously profitable business
even if the house is honest and even if the house waives the modicum, the
dead spot on the roulette wheel, that is its recognized due. The reason is
that the addicted gambler, in his greed or desperation, will persist in
playing until he has lost everything, and this will happen before the house
has exhausted its ample reserves. It is a profitable business and a
heartless one."

Now that's quite wrong. If a casino ran nothing but fair games with no edge
to the house, then it would roughly break even, no matter how insanely hard
its customers tried to work at losing all their money. Still, the idea seems
"reasonable," does it not?

The paragraph I quoted was written by one of the most eminent mathematical
logicians of the 20th century, who surely should have known better.

--walter trice

### Douglas Zare

Dec 23, 1999, 3:00:00 AM12/23/99
to
Everything was fine until Douglas Zare wrote:

> [...]

> On the other hand, the two situations Walter Trice analyzed initially give
> quite different answers to dmg's question about the bias toward 6's. If all
> checkers are on the ace point there is no bias toward 6's, but with one

> checker there will be a 6/(1+2+...+6)=2/7 bias again. [...]

That would be if one were rolling the dice one at a time. Oops. (Thanks to Walter
Trice for correcting me by e-mail.) The 11 of the 36 rolls with 6's give 114 of
294 pips, so the probability of a single checker far away coming home on a roll
containing a 6 is 114/294 ~38.8% rather than the 28.6% (24 of 84 numbers) one
gets for checkers piled on the ace point.

The latter is slightly lower than the 11/36 chance of rolling a 6. Under the
condition that one rolls at least one 6, the chance of rolling double 6's is
1/11, lower than the 1/5 chance of rolling doubles under the condition that one
does not roll a 6.

For many positions, the bias toward 6's will be greater than either of these. If
both sides have a checker on the 6 and 1, then assuming proper play the winning
side will have rolled a 6
11/36 *(1+21/36 + (21/36)^2+(2*2*21)/(36^3)+(2/36)^4)~58.8% of the time, assuming
that I have done these calculations correctly. (The five terms correspond to the
5 possible ending times, with the last possible only after 4 1-2's.)

Douglas Zare

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